Question

Вычислите интеграл методом интегрирования по частям.

Answer 1

Решение:

$\int\limits^2_0 {(4-x)*7^{x}} \, dx=\int\limits {(4-x)*7^{x}} \, dx=\int\limits {4*7^{x}-x*7^{x}} \, dx=\int\limits {4*7^{x}} \, dx-\int\limits {x*7^{x}} \, dx=4*\int\limits {7^{x}} \, dx-\int\limits {x*7^{x}} \, dx=4*\frac{7^{x}}{ln7}-(x*\frac{7x^{x}}{ln7}-\int\limits {\frac{7^{x}}{ln7}} \, dx)=4*\frac{7^{x}}{ln7}-(x*\frac{7x^{x}}{ln7}-\frac{1}{ln7}*\int\limits {7^{x}} \, dx)=4*\frac{7^{x}}{ln7}-(x*\frac{7x^{x}}{ln7}-\frac{1}{ln7}*\frac{7^{x}}{ln7})=$$=4*\frac{7^{x}}{ln7}-(\frac{x*7^{x}}{ln7}-\frac{7^{x}}{ln^{2}7})=4*\frac{7^{x}}{ln7}-\frac{x*7^{x}}{ln7}+\frac{7^{x}}{ln^{2}7}=(\frac{4*7^{x}-x*7^{x}}{ln7}+\frac{7^{x}}{ln^{2}7})|^{2}_{0}=\frac{4*7^{2}-2*7^{2}}{ln7}+\frac{7^{2}}{ln^{2}7}-(\frac{4*7^{0}-0*7^{0}}{ln7}+\frac{7^{0}}{ln^{2}7})=\frac{4*49-2*49}{ln7}+\frac{49}{ln^{2}7}-(\frac{4*1-0*1}{ln7}+\frac{1}{ln^{2}7})=\frac{196-98}{ln7}+\frac{49}{ln^{2}7}-(\frac{4-0}{ln7}+\frac{1}{ln^{2}7})=$

$=\frac{98}{ln7}+\frac{49}{ln^{2}7}-(\frac{4}{ln7}+\frac{1}{ln^{2}7})=\frac{98}{ln7}+\frac{49}{ln^{2}7}-\frac{4}{ln7}-\frac{1}{ln^{2}7}=\frac{94}{ln7}+\frac{48}{ln^{2}7}$

Ответ: $\frac{94}{ln7}+\frac{48}{ln^{2}7}$

Answer 2

Ответ:

$\int\limits^2_0\, \underbrace{(4-x)}_{u}\cdot \underbrace{7^{x}\, dx}_{dv}=\Big[\ du=-dx\ ,\ v=\dfrac{7^{x}}{ln7}\ \Big]=uv-\int v\, du=\\\\\\=(4-x)\cdot \dfrac{7^{x}}{ln7}\Big|_0^2+\dfrac{1}{ln7}\cdot \int \limits _0^27^{x}\, dx=(4-x)\cdot \dfrac{7^{x}}{ln7}\Big|_0^2+\dfrac{7^{x}}{ln^27}\Big|_0^2=\\\\\\=\dfrac{1}{ln7}\cdot (2\cdot7^2-4\cdot 7^0)+\dfrac{1}{ln^27}\cdot (7^2-7^0)=\\\\\\=\dfrac{1}{ln7}\cdot (98-4)+\dfrac{1}{ln^27}\cdot (49-1)=\dfrac{94}{ln7}+\dfrac{48}{ln^27}$